3.27.17 \(\int \frac {3+(1-2 k^2) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{(1-x^2) (1-k^2 x^2)} (-1+d-(1+2 d) x+(d+k^2) x^2+k^2 x^3)} \, dx\) [2617]

Optimal. Leaf size=229 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2 \sqrt [3]{d}-2 \sqrt [3]{d} x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{\sqrt [3]{d}}+\frac {\log \left (d^{2/3}-2 d^{2/3} x+d^{2/3} x^2+\left (\sqrt [3]{d}-\sqrt [3]{d} x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]

[Out]

-3^(1/2)*arctan(3^(1/2)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/3)/(2*d^(1/3)-2*d^(1/3)*x+(1+(-k^2-1)*x^2+k^2*x^4)^(1/3)))
/d^(1/3)-ln(-d^(1/3)+d^(1/3)*x+(1+(-k^2-1)*x^2+k^2*x^4)^(1/3))/d^(1/3)+1/2*ln(d^(2/3)-2*d^(2/3)*x+d^(2/3)*x^2+
(d^(1/3)-d^(1/3)*x)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/3)+(1+(-k^2-1)*x^2+k^2*x^4)^(2/3))/d^(1/3)

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Rubi [F]
time = 1.55, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3 + (1 - 2*k^2)*x - 3*k^2*x^2 + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(-1 + d - (1 + 2*d)*x + (d + k^
2)*x^2 + k^2*x^3)),x]

[Out]

(x*(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*AppellF1[1/2, 1/3, 1/3, 3/2, x^2, k^2*x^2])/(1 - (1 + k^2)*x^2 + k^2*x^
4)^(1/3) - (4 - d)*Defer[Int][1/((1 - d + (1 + 2*d)*x - (d + k^2)*x^2 - k^2*x^3)*(1 + (-1 - k^2)*x^2 + k^2*x^4
)^(1/3)), x] - 2*(1 + d - k^2)*Defer[Int][x/((1 - d + (1 + 2*d)*x - (d + k^2)*x^2 - k^2*x^3)*(1 + (-1 - k^2)*x
^2 + k^2*x^4)^(1/3)), x] + (d + 4*k^2)*Defer[Int][x^2/((1 - d + (1 + 2*d)*x - (d + k^2)*x^2 - k^2*x^3)*(1 + (-
1 - k^2)*x^2 + k^2*x^4)^(1/3)), x]

Rubi steps

\begin {align*} \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}}+\frac {4-d+2 \left (1+d-k^2\right ) x-\left (d+4 k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {4-d+2 \left (1+d-k^2\right ) x-\left (d+4 k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {\left (1-\frac {4}{d}\right ) d}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )}+\frac {2 \left (-1-d+k^2\right ) x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )}+\frac {\left (d+4 k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left ((-4+d) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (2 \left (1+d-k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\left (d+4 k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d+(1+2 d) x-\left (d+k^2\right ) x^2-k^2 x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [F]
time = 41.41, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3+\left (1-2 k^2\right ) x-3 k^2 x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d-(1+2 d) x+\left (d+k^2\right ) x^2+k^2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(3 + (1 - 2*k^2)*x - 3*k^2*x^2 + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(-1 + d - (1 + 2*d)*x + (
d + k^2)*x^2 + k^2*x^3)),x]

[Out]

Integrate[(3 + (1 - 2*k^2)*x - 3*k^2*x^2 + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(-1 + d - (1 + 2*d)*x + (
d + k^2)*x^2 + k^2*x^3)), x]

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {3+\left (-2 k^{2}+1\right ) x -3 k^{2} x^{2}+k^{2} x^{3}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{3}} \left (-1+d -\left (1+2 d \right ) x +\left (k^{2}+d \right ) x^{2}+k^{2} x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3+(-2*k^2+1)*x-3*k^2*x^2+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d-(1+2*d)*x+(k^2+d)*x^2+k^2*x^3),x)

[Out]

int((3+(-2*k^2+1)*x-3*k^2*x^2+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d-(1+2*d)*x+(k^2+d)*x^2+k^2*x^3),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+(-2*k^2+1)*x-3*k^2*x^2+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d-(1+2*d)*x+(k^2+d)*x^2+k^2*x^3
),x, algorithm="maxima")

[Out]

integrate((k^2*x^3 - 3*k^2*x^2 - (2*k^2 - 1)*x + 3)/((k^2*x^3 + (k^2 + d)*x^2 - (2*d + 1)*x + d - 1)*((k^2*x^2
- 1)*(x^2 - 1))^(1/3)), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+(-2*k^2+1)*x-3*k^2*x^2+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d-(1+2*d)*x+(k^2+d)*x^2+k^2*x^3
),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+(-2*k**2+1)*x-3*k**2*x**2+k**2*x**3)/((-x**2+1)*(-k**2*x**2+1))**(1/3)/(-1+d-(1+2*d)*x+(k**2+d)*x
**2+k**2*x**3),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+(-2*k^2+1)*x-3*k^2*x^2+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d-(1+2*d)*x+(k^2+d)*x^2+k^2*x^3
),x, algorithm="giac")

[Out]

integrate((k^2*x^3 - 3*k^2*x^2 - (2*k^2 - 1)*x + 3)/((k^2*x^3 + (k^2 + d)*x^2 - (2*d + 1)*x + d - 1)*((k^2*x^2
- 1)*(x^2 - 1))^(1/3)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {3\,k^2\,x^2-k^2\,x^3+x\,\left (2\,k^2-1\right )-3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/3}\,\left (d+k^2\,x^3+x^2\,\left (k^2+d\right )-x\,\left (2\,d+1\right )-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*k^2*x^2 - k^2*x^3 + x*(2*k^2 - 1) - 3)/(((x^2 - 1)*(k^2*x^2 - 1))^(1/3)*(d + k^2*x^3 + x^2*(d + k^2) -
x*(2*d + 1) - 1)),x)

[Out]

-int((3*k^2*x^2 - k^2*x^3 + x*(2*k^2 - 1) - 3)/(((x^2 - 1)*(k^2*x^2 - 1))^(1/3)*(d + k^2*x^3 + x^2*(d + k^2) -
x*(2*d + 1) - 1)), x)

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